calculate Rx,R1,R2,R3

 

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i started by removing the the branch for R1,R2 and R3 and short RX found the current for the circuit ..the trouble is that when i took back R1,R2 and R3,i found that i do not have the voltage ranges in order for me to calculate the resistors.This is the area i need help,may there other ways of solving that i don't know.

 

Are you trying calculate or measure?

R1, R2, R3 is to bypass excess current to the ammeter. Values of these depends on the battery voltage.

If you place any these bypass resistors in parallel with the ammeter, then the potentiometer has to be adjusted (with Rx shorted) so that the maximum ammeter current is achieved. Once that is done, the meter branch with Rm and Rx becomes a basic voltage divider/current limiter circuit with a fairly constant voltage (Vm = Im x Rm = 2000 Im) across the parallel circuit; assuming R1 + R2 + R3 < 200 ohms. [Rule of thumb : if the total bypass resistance of a parallel ciruit is one-tenth of the other other branch, current through the other branch will be insignificant, therefore voltage drop can be considered as that of current through the bypass resistors multiplied by the bypass resistance]

As such, with Rx in circuit, Vm = Im' x (Rx + Rm) = Im' x (Rx + 2000). If you are measuring, then Im' can be read from the meter, the only unknown will be Rx and that can be calculated -> Rx = (Vm/Im') - 2000

Back to R1, R2, R3. To choose these resistances:
The two extreme conditions : 0 ohm < Rpot < 5000 ohms
Let take Rpot = 5000 ohhms and Vbatt = 1.5V with Rx = 0 ohm (shorted) and Im = maximum meter current
** Vm = Im x 2000 = I x (R1 + R2 + R3)
Using the one-tenth Rule of Thumb, we can ignore the effects of Im,
I x (Rpot) + Vm = Vbatt
5000 I + Vm = 1.5
I = (1.5 - Vm)/5000

Substituting this back into **
R1 + R2 + R3 = Vm/I = 5000 Vm/ (1.5 - Vm)

Typically Vm is around 50 microvolts, as suchthe denominator can be estimated as 1.5:
R1 + R2 + R3 = 5000 Vm/1.5 = 3333 Vm

 

Are you trying calculate or measure?

R1, R2, R3 is to bypass excess current to the ammeter. Values of these depends on the battery voltage.

If you place any these bypass resistors in parallel with the ammeter, then the potentiometer has to be adjusted (with Rx shorted) so that the maximum ammeter current is achieved. Once that is done, the meter branch with Rm and Rx becomes a basic voltage divider/current limiter circuit with a fairly constant voltage (Vm = Im x Rm = 2000 Im) across the parallel circuit; assuming R1 + R2 + R3 < 200 ohms. [Rule of thumb : if the total bypass resistance of a parallel ciruit is one-tenth of the other other branch, current through the other branch will be insignificant, therefore voltage drop can be considered as that of current through the bypass resistors multiplied by the bypass resistance]

As such, with Rx in circuit, Vm = Im' x (Rx + Rm) = Im' x (Rx + 2000). If you are measuring, then Im' can be read from the meter, the only unknown will be Rx and that can be calculated -> Rx = (Vm/Im') - 2000

Back to R1, R2, R3. To choose these resistances:
The two extreme conditions : 0 ohm < Rpot < 5000 ohms
Let take Rpot = 5000 ohhms and Vbatt = 1.5V with Rx = 0 ohm (shorted) and Im = maximum meter current
** Vm = Im x 2000 = I x (R1 + R2 + R3)
Using the one-tenth Rule of Thumb, we can ignore the effects of Im,
I x (Rpot) + Vm = Vbatt
5000 I + Vm = 1.5
I = (1.5 - Vm)/5000

Substituting this back into **
R1 + R2 + R3 = Vm/I = 5000 Vm/ (1.5 - Vm)

Typically Vm is around 50 microvolts, as suchthe denominator can be estimated as 1.5:
R1 + R2 + R3 = 5000 Vm/1.5 = 3333 Vm

 

Thank u got the point